Solve the equation that is in quadratic form.

(2x+5)2+21=−10(2x+5)

Question

Solve the equation that is in quadratic form.

(2x+5)2+21=−10(2x+5)

BlackTom AI · Accepted Answer

The problem asks us to solve a quadratic equation written in a disguised form. We can simplify the process by introducing a substitution to reduce the equation to a standard quadratic.
First, let y = 2x + 5. The equation becomes y^2 + 21 = -10y, which can be rearranged to bring all terms to one side: y^2 + 10y + 21 = 0.
Now look for two numbers that multiply to 21 and add to 10. Those numbers are 3 and 7, but since the middle term is positive, we want factors (y + 3)(y + 7) = 0.
From this factorization, we get two possible values for y: y = -3 or y = -7.
Proceed to revert back to x using y = 2x + 5.
Case 1: y = -3 implies 2x + 5 = -3, so 2x = -8 and x = -4.
Case 2: y = -7 implies 2x + 5 = -7, so 2x = -12 and x = -6.
Therefore, the solution set is x = -4 and x = -6, which can be written in either order as -4, -6 or -6, -4. 
If you compare the two potential answer arrangements, both express the same pair of solutions, just listed in different sequences. Neither option is incorrect as long as both roots are included. 
A quick check: substitute x = -4 into the original equation: (2(-4)+5)^2 + 21 equals (-3)^2 + 21 equals 9 + 21 = 30, and -10(2(-4)+5) equals -10(-3) equals 30, so x = -4 satisfies the equation. Substituting x = -6 yields (2(-6)+5)^2 + 21 equals (-7)^2 + 21 equals 49 + 21 = 70, and -10(2(-6)+5) equals -10(-7) equals 70, so x = -6 also satisfies the equation. 
Thus, the solution set is {-6, -4} (in any order).

MATH136-12P-IO-202510-I-81X M04: Midterm Exam Review Quiz - Timed

Solve the equation that is in quadratic form. (2x+5)2+21=−10(2x+5)

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